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We have a smooth embedded submanifold $M\subseteq\mathbb{R}^n$ and we want to show that there is a $c\in \mathbb{R}$ such that the plane $x_1=c$ intersects $M$ transversally. Show that the set of $c$ is not necessarily open in $\mathbb{R}$.

I have beat my head against this problem, but while I know the definition of intersecting transversally (submanifolds of $Y,$ $M$ and $N$ intersect transversally if $\forall x\in M\cap N,$ $T_x M+T_x N=T_x Y$) I really struggle to work with this definition.

Cameron
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    Here is an idea for show that the set of $c$ is not open : take $C$ the graph of a function $f : \Bbb R \to \Bbb R$ such that the local minimums of the function are exactly the set $ {n}_{n \in \Bbb N}$, and $f(n) = 1/n$. Then, $x = 0$ is transverse to $C$ but there is no $\varepsilon > 0$ such that for all $c \in (-\varepsilon, \varepsilon)$, $x=c$ is transverse to the curve. – Nicolas Hemelsoet Dec 15 '17 at 07:44

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